Find the value of $\sin^{-1}(\cos 2)-\cos^{-1}(\sin 2) +\tan^{-1}(\cot 4) -\cot^{-1}(\tan 4)+\sec^{-1}(\csc 6)-\csc^{-1}(\sec 6)$ The given expression simplifies to
$$\sin^{-1}(\sin 2)-\cos^{-1}(\cos 2)+\tan^{-1}(\tan 4)-\cot^{-1}(\cot 4)+\sec^{-1}(\sec 6)-\csc^{-1} (\csc 6)$$
$$=(\pi-2)-2+(4-\pi)-(2\pi-4)+(2\pi-6)-(2\pi-6)$$
$$=-2\pi+4$$
But the given answer is $5\pi-16$. I rechecked all the the principle branches, and they all seem to be right. Where did use the wrong value?
 A: These type of questions are very easy to handle by graphs. Just remember their graphs and for finding the value at each point, see the location of that point and write the equation of the line by seeing it's slope and the point where it cuts the $x$ axis and you will get the value. For example, you have to find $\csc^{-1}(\csc 6)$, in it's graph note that $6$ lies in between $\frac{3\pi}{2}$ and $2\pi$ and the line has a positive slope and the cutting point with $x$ axis is $x=2\pi$ and hence the equation of the line would be $y=x-2\pi$.
So your mistakes were $\cot^{-1}(\cot 4)=\pi-4$,  not $2\pi-4$, and secondly $\csc^{-1}(\csc 6)=6-2\pi$ and not $2\pi-6$. Hence the correct answer will come out to be $5\pi-16$.
A: I don't get your simplification. How did $\sin^{-1}(\cos 2) - \cos^{-1}(\sin 2)$ become $\sin^{-1}(\sin 2) - \cos^{-1}(\cos 2)$? I think the simplification part is off.

We must identify which quadrant each of these lie in. For example, $2$ lies in the second quadrant, because $\pi > 2 > \frac \pi 2$. $4$ lies in the third quadrant, and $6$ lies in the fourth quadrant.
With this in mind, $\sin^{-1}(\cos 2) = \frac{\pi}{2} - 2$. Similarly, $\cos^{-1}(\sin 2) = 2 - \frac \pi 2$, because the inverse cosine function is restricted to $(-\frac{\pi}{2},\frac{\pi}{2}]$.

*

*$\tan^{-1}(\cot 4) = \frac{3 \pi}{2} - 4$.


*$\cot^{-1} (\tan 4) = \frac{3 \pi }{2} - 4$.


*$\sec^{-1}(\csc 6) = \frac{5 \pi}{2} - 6$.


*$\csc^{-1}(\sec 6) = 6 - \frac{3 \pi}{2}$.
(You can convince yourself of these and finish the problem).

With the simplified expression :

*

*$\sin^{-1}(\sin 2) = \pi- 2$


*$\cos^{-1}(\cos 2) = 2$


*$\tan^{-1}(\tan 4) = 4-\pi$.


*$\color{green}{\cot^{-1}(\cot 4) = 4-\pi}$. Explanation : the principal region for $\cot^{-1}$ is $[-\frac \pi 2, \frac \pi 2]$, and $\cot 4$ is positive because $4$ is in the third quadrant. Thus, $\cot^{-1}(\cot 4)$ lies in the first quadrant, so the angle is positive, hence $4-\pi$ (and not $\pi -4$, which is a negative angle).


*$\sec^{-1}(\sec 6) = 2\pi - 6$.


*$\color{green}{\csc^{-1}(\csc 6) = 6-2\pi}$. Explanation : the principal region of $\csc^{-1}$ is $[-\frac {\pi}{2} , \frac{\pi}{2}]$ , and  $6$ belongs in the fourth quadrant, so is already in the prescribed region, as long as we subtract $2\pi$, which gives $6 - 2\pi$ (and since the cosecants of both these angles are negative, this makes sense).
Now evaluating gives the right answer.
